Problem: $9pq - 6pr - 2p - 9 = q + 4$ Solve for $p$.
Combine constant terms on the right. $9pq - 6pr - 2p - {9} = q + {4}$ $9pq - 6pr - 2p = q + {13}$ Notice that all the terms on the left-hand side of the equation have $p$ in them. $9{p}q - 6{p}r - 2{p} = q + 13$ Factor out the $p$ ${p} \cdot \left( 9q - 6r - 2 \right) = q + 13$ Isolate the $p$ $p \cdot \left( {9q - 6r - 2} \right) = q + 13$ $p = \dfrac{ q + 13 }{ {9q - 6r - 2} }$